Let \(R\) be a PID, let \(I\subsetneq R\) be a proper ideal, and suppose \(a\in I\) is irreducible. Prove that \[ I=(a). \]
Solution
Since \(R\) is a PID, write \(I=(d)\). Because \(a\in I\), there exists \(r\in R\) with \[ a=dr. \] As \(a\) is irreducible, either \(d\) is a unit or \(r\) is a unit. The first possibility would give \(I=(d)=R\), contradicting that \(I\) is proper. Hence \(r\) is a unit, so \(d\) and \(a\) are associates. Therefore \[ I=(d)=(a). \]